GẤP GẤP, GIÚP TUI CÂU NÀY VỚI CÁC BẠN ƠI
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b: Tọa độ giao điểm là:
\(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=-\dfrac{3}{2}x-1\\y=\dfrac{1}{2}x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\y=\dfrac{1}{2}x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-4\end{matrix}\right.\)
\(c,\overline{735a2b}⋮̸2;⋮5\Rightarrow b=5\\ \Rightarrow\overline{735a25}⋮9\\ \Rightarrow7+3+5+a+2+5⋮9\\ \Rightarrow22+a⋮9\\ \Rightarrow a=5\\ \Rightarrow\overline{735a2b}=735525\\ b,\overline{5a27b}⋮2;⋮5\Rightarrow b=0\\ \Rightarrow\overline{5a270}⋮9\\ \Rightarrow5+a+2+7+0⋮9\\ \Rightarrow14+a⋮9\\ \Rightarrow a=4\\ \Rightarrow\overline{5a27b}=54270\)
\(4^{x+2}+4^x=272\)
\(\Leftrightarrow4^x=16\)
hay x=2
quy đồng mẫu số : 1/4=7/28 ; 2/7= 8/28
Vì 8/18 > 7/28 nên 2/4 > 1/7
Vậy Hòa được tặng nhiều hoa hơn
\(t,=\dfrac{3^{64}\cdot7^{24}}{7^{23}\cdot9^{66}}=\dfrac{7}{3^2}=\dfrac{7}{9}\\ u,=\dfrac{5^3\cdot3^4\cdot2^5}{5^2\cdot3^3\cdot2^4}=5\cdot3\cdot2=30\\ v,=\dfrac{3^6\cdot2^{15}}{2^6\cdot3^6\cdot2^8}=2\)
a: 6/5-2/3=18/15-10/15=8/15
b: 9/7-5/8=72/56-35/56=37/56
c: 7/6-3/4=14/12-9/12=5/12
d: 9/8-8/9=81/72-64/72=17/72
e: 7/5-7/6=7(1/5-1/6)=7/30
f: 3/4-2/5=15/20-8/20=7/20
g: 4/7-5/6=24/42-35/42=-11/41
i: 6/13-5/11=66/143-65/143=1/143
k: 1-4/5=1/5
m: 2-5/9=18/9-5/9=13/9
n: 3-9/5=15/5-9/5=6/5
p: 4-3/4=16/4-3/4=13/4
a) \(\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18}{15}-\dfrac{10}{15}=\dfrac{8}{15}\)
b) \(\dfrac{9}{7}-\dfrac{5}{8}=\dfrac{72}{56}-\dfrac{35}{56}=\dfrac{37}{56}\)
c) \(\dfrac{7}{6}-\dfrac{3}{4}=\dfrac{28}{24}-\dfrac{18}{24}=\dfrac{10}{24}=\dfrac{5}{12}\)
d) \(\dfrac{9}{8}-\dfrac{8}{9}=\dfrac{81}{72}-\dfrac{64}{72}=\dfrac{17}{72}\)
e) \(\dfrac{7}{5}-\dfrac{7}{6}=\dfrac{42}{30}-\dfrac{35}{30}=\dfrac{7}{30}\)
h) \(\dfrac{4}{7}-\dfrac{5}{6}=\dfrac{24}{42}-\dfrac{35}{42}=-\dfrac{11}{42}\)
i) \(\dfrac{6}{13}-\dfrac{5}{11}=\dfrac{66}{143}-\dfrac{65}{143}=\dfrac{1}{143}\)
k) \(1-\dfrac{4}{5}=\dfrac{5}{5}-\dfrac{4}{5}=\dfrac{1}{5}\)
m) \(2-\dfrac{5}{9}=\dfrac{18}{9}-\dfrac{5}{9}=\dfrac{13}{9}\)
n) \(3-\dfrac{9}{5}=\dfrac{15}{5}-\dfrac{9}{5}=\dfrac{6}{5}\)
p) \(4-\dfrac{3}{4}=\dfrac{16}{4}-\dfrac{3}{4}=\dfrac{13}{4}\)
7: \(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
=>300-x=0
hay x=300
\(P=\left(x^2+2x\right)\left(y^2-4y\right)+5\left(x^2+2x\right)+4\left(y^2-4y\right)+2021\)
\(=\left[\left(x+1\right)^2-1\right]\left[\left(y-2\right)^2-4\right]+5\left(x+1\right)^2+4\left(y-2\right)^2+2000\)
\(=\left(x+1\right)^2\left(y-2\right)^2+\left(x+1\right)^2+3\left(y-2\right)^2+2024\ge2024\)
\(P_{min}=2024\) khi \(\left(x;y\right)=\left(-1;2\right)\)